Note: This questions is quite different from the ones referred to in the comments. Those deal with numerical questions, while this one is algebraic.

I have plots of the following type:

Plot[Cos[50 t] + Cos[51 t], {t, 0, 10}]

enter image description here

I would like to plot a envelope over this plot, i.e. another plot that joins all of maxima and minima of this plot respectively. Here is my attempt, but it's not exactly what I'd like:

Plot[{Cos[50 t] + Cos[51 t], Cos[t] + 1.5, -Cos[t] - 1.5}, {t, 0, 10}]

enter image description here

How can I generate the actual envelope?

  • $\begingroup$ I'm guessing that's something like Cos[42x]+Cos[43x]? $\endgroup$ Jun 3, 2014 at 18:29
  • $\begingroup$ this is completely true but this is for the Fig.1, I want to access to sheath while I do not access to any formula for that. $\endgroup$ Jun 3, 2014 at 18:32
  • $\begingroup$ Well, you've got to give us some kind of input to start with. $\endgroup$ Jun 3, 2014 at 18:34
  • $\begingroup$ I plotted these ones with: Plot[{Cos[50 t] + Cos[51 t], Cos[t] + 1.5, -Cos[t] - 1.5}, {t, 0, 10}], I used of simulated functions 'Cos[t] + 1.5' and '-Cos[t] - 1.5' for the sheath. $\endgroup$ Jun 3, 2014 at 18:36
  • 1
    $\begingroup$ Nearly a duplicate of Elegant way of obtaining the envelope of oscillating function, which is a duplicate of Mathematica envelope for the bottom of a plot, a generic function. But this one just straight trigonometry. $\endgroup$
    – Michael E2
    Aug 28, 2014 at 0:20

2 Answers 2


Playing with the manipulate below might help. It's based on the the acoustics of beats.

  {Cos[a*t] + Cos[b*t], 2*Cos[(b - a) t/2], -2*Cos[(b - a) t/2]}, {t, 0, 10},
  PlotStyle -> {
   Directive[Black, Thick], 
   Directive[Black, Thick]}],
 {{a, 20}, 1, 50}, {{b, 21}, 1, 50}]

enter image description here

  • $\begingroup$ Thank you so much. this is correct $\endgroup$ Jun 3, 2014 at 18:58

Don't mind me, I'm just having fun.

Grab the definition of HilbertTransform from this previous post, and then:

f[t_] := Cos[50 t] + Cos[51 t] + Sin[53 t] (* more sinusoids = more fun *)
g[t_] := Evaluate@HilbertTransform[f[τ], τ, t]
h[t_] := Abs[f[t] + I g[t]]
Plot[{f[t], h[t], -h[t]}, {t, 0, 10}, ImageSize -> Large, PlotPoints -> 100,
 PlotStyle -> {Automatic, Black, Black}]

enter image description here

You can see that the envelope has a nice analytical form:

ComplexExpand[h[t]] // FullSimplify

$\sqrt{3 + 2\cos t + 2\sin 2t + 2\sin 3t}$

Further reading: analytic representation.

  • $\begingroup$ This way though doesn't seem to catch when the envelope should go to zero. Shouldn't it e.g. vanish near 2? $\endgroup$
    – Ruslan
    Jun 4, 2014 at 8:34
  • 1
    $\begingroup$ Should it? It still looks like a sinusoid with small but nonzero amplitude there. (The envelope correctly goes to zero for the original function in the question, if that's what you're concerned about.) $\endgroup$
    – user484
    Jun 4, 2014 at 8:55
  • $\begingroup$ Ah, OK then, you're right. $\endgroup$
    – Ruslan
    Jun 4, 2014 at 9:02
  • $\begingroup$ @Rahul: wow! that's really cool! Would this also work on a set of discrete data? Something like ListLinePlot[Accumulate[RandomReal[{-1, 1}, 1000]]] ... If not, how could one modify it so that it does? I have a bunch of data that makes a seemingly random curve like that and I'd like be able to integrate over some area under the curve like the one that you generated. (Hmmm, i just copy pasted your code, but didn't get the black curve :( what could have gone wrong?) $\endgroup$
    – Raksha
    May 16, 2015 at 3:46
  • 1
    $\begingroup$ @Solarmew: You first need to copy the definition of the Hilbert transform from the post I linked to. It also gives a discrete version of the transform, which you could use in much the same way on discrete data. $\endgroup$
    – user484
    May 16, 2015 at 15:58

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